// 122. Best Time to Buy and Sell Stock II
//
// Say you have an array for which the ith element is the price of a given stock on day i.
//
// Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).
//
// Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).
//
// Example 1:
//
// Input: [7,1,5,3,6,4]
// Output: 7
// Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
// Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
//
// Example 2:
//
// Input: [1,2,3,4,5]
// Output: 4
// Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
// Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
// engaging multiple transactions at the same time. You must sell before buying again.
//
// Example 3:
//
// Input: [7,6,4,3,1]
// Output: 0
// Explanation: In this case, no transaction is done, i.e. max profit = 0.
//
// https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/

/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function (prices) {
	let profit = 0
	let buyPrice = -1
	let len = prices.length

	prices.forEach((currentPrice, index) => {
		let nextPrice = (index + 1 < len) ? prices[index + 1] : 0

		if (buyPrice === -1) {
			buyPrice = currentPrice
		} else {
			if (currentPrice > buyPrice) {
				if (nextPrice < currentPrice) {
					profit += (currentPrice - buyPrice)
					buyPrice = -1
				}
			} else {
				buyPrice = currentPrice
			}
		}
	})

	return profit
};

console.log(maxProfit([7,1,5,3,6,4]))
console.log(maxProfit([1,2,3,4,5]))
console.log(maxProfit([7,6,4,3,1]))